1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Help: Vector field and radius vector

  1. Oct 1, 2006 #1
    Hi Guys,

    Given the vector field [tex]X(x,y) = ( a + \frac{b(y^2-x^2)}{(x^2+y^2)^2}, \frac{-2bxy}{(x^2+y^2)^2}}})[/tex]

    Show that for a point (x,y) on the circle with radius r = \sqrt(b/a) (i.e. x^2 + y^2 = b/a), the vector X(x,y) is tangent to a circle at the point.

    My strategy is that to first define a vector function r(t) and show that this function is to a point on the circle?

    Cheers and Best Whishes MM23

    p.s. the potential function for X(x,y) is F(x,y) = ax + (bx/x^2+y^2).

    p.p.s. Do I first find the directional dereative for F(x,y) ?

    And then check to see if this vector is a tangent for the circle?

    my solution:


    [tex]r (\theta )=\sqrt{\tfrac{b}{a}}(\cos \theta , \, \sin \theta ), \qquad 0\le \theta <2\pi [/tex]

    be any point on the circle. Plugging into the vector field one get at such a point the vector

    [tex]$a(1+\sin ^2 \theta -\cos ^2 \theta ,\, -2\cos \theta \sin \theta )=2a (\sin ^2 \theta ,\, -\cos \theta \sin \theta ).[/tex]

    One realizes upon a scalar multiplication by the radius vector that this vector is orthogonal to it. Thus...

    If this is correct to I then conclude that [tex]r (\theta ) \cdot X = 0 [/tex] ??
    Last edited: Oct 1, 2006
  2. jcsd
  3. Oct 1, 2006 #2


    User Avatar
    Science Advisor

    Yes, that works nicely.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook